nand2tetris_ALU

这一节,我们将尝试构建CPU中的ALU单元。明明上一节才开始学习基本逻辑门,这一节就实现ALU,当时的我是吃惊的,但 确实仅用逻辑门就可以完成。在开始逐步实现之前,先补充一些前置知识

前置知识

HDL

上一节构建选择器时,得到了一个较长的函数式,那么如何验证函数表达式呢。课程老师提供了用于描述电路的HDL规范和解释运行的硬件模拟器,这些都在课程软件包中(之前实现MIT 6.824时,也是有各种模板代码和测试脚本,再次忍不住对Top大学的心向往之),执行示例图如下

HDL引入

先来看一下之前的选择器的HDL实现,简单介绍一下HDL规范

// This file is part of www.nand2tetris.org
// and the book "The Elements of Computing Systems"
// by Nisan and Schocken, MIT Press.
// File name: projects/1/Mux.hdl
/** 
 * Multiplexor:
 * if (sel = 0) out = a, else out = b
 */
CHIP Mux {
    IN a, b, sel;
    OUT out;

    PARTS:
    Not (in = sel, out = Notsel);
    And (a = a, b = Notsel, out = aAndNotsel);
    And (a = b, b = sel, out = bAndsel);
    Or (a = aAndNotsel, b = bAndsel, out = out);
}

IN 定义输入形参名,OUT 定义输出形参名;Not可以理解为函数,in、a、b和out为函数参数名,sel和Notsel为函数传参;还有许多需要注意和特殊的语法的地方,一方面查看模拟器报错信息,另一方面参考老师标准实现,不建议在这里花费太多时间,毕竟我们只是学习思路
下面再介绍几个有趣的HDL实现,更好的感受HDL规范,对了,还有一点,HDL是电路图描述语言,如果出现电路回环,完全可以先使用变量,再声明(这一点,可能要在第三节寄存器实现)

有趣的HDL实现

多函数调用

/**
 * Exclusive-or gate:
 * if ((a and Not(b)) or (Not(a) and b)) out = 1, else out = 0
 */
CHIP Xor {
    IN a, b;
    OUT out;

    PARTS:
    Not (in = a, out = Nota);
    Not (in = b, out = Notb);
    
    And (a = a, b = Notb, out = aAndNotb);
    And (a = b, b = Nota, out = bAndNota);
    
    Or (a = aAndNotb, b = bAndNota, out = out);
}

多路总线的使用

/**
 * 16-bit multiplexor: 
 * for i = 0, ..., 15:
 * if (sel = 0) out[i] = a[i], else out[i] = b[i]
 */
CHIP Mux16 {
    IN a[16], b[16], sel;
    OUT out[16];

    PARTS:
    Mux (a = a[0], b = b[0], sel = sel, out = out[0]);
    Mux (a = a[1], b = b[1], sel = sel, out = out[1]);
    Mux (a = a[2], b = b[2], sel = sel, out = out[2]);
    Mux (a = a[3], b = b[3], sel = sel, out = out[3]);
    Mux (a = a[4], b = b[4], sel = sel, out = out[4]);
    Mux (a = a[5], b = b[5], sel = sel, out = out[5]);
    Mux (a = a[6], b = b[6], sel = sel, out = out[6]);
    Mux (a = a[7], b = b[7], sel = sel, out = out[7]);
    Mux (a = a[8], b = b[8], sel = sel, out = out[8]);
    Mux (a = a[9], b = b[9], sel = sel, out = out[9]);
    Mux (a = a[10], b = b[10], sel = sel, out = out[10]);
    Mux (a = a[11], b = b[11], sel = sel, out = out[11]);
    Mux (a = a[12], b = b[12], sel = sel, out = out[12]);
    Mux (a = a[13], b = b[13], sel = sel, out = out[13]);
    Mux (a = a[14], b = b[14], sel = sel, out = out[14]);
    Mux (a = a[15], b = b[15], sel = sel, out = out[15]);
}

组合已有实现,实现更高级功能

/**
 * 4-way 16-bit multiplexor:
 * out = a if sel = 00
 *       b if sel = 01
 *       c if sel = 10
 *       d if sel = 11
 */
CHIP Mux4Way16 {
    IN a[16], b[16], c[16], d[16], sel[2];
    OUT out[16];
    
    PARTS:
    Mux16 (a = a, b = b, sel = sel[0], out = mux1);
    Mux16 (a = c, b = d, sel = sel[0], out = mux2);
    Mux16 (a = mux1, b = mux2, sel = sel[1], out = out);
}

二进制数

十进制数0-9,二进制数0-1,大于最大数时则进位。基本的二进制数容易理解,这里单独列出来,主要想讲一下负数,然后引出减法实现,毕竟一个ALU你不能算减法这太糟糕了,至于乘法和除法,则使用软件来实现吧

负数

一个四位的数,0000 - 1111,如果全部用来表示正数的话 0 - 15,这也被称为无符号表示。若想表示负数,取第一位做标识符,0为正,1为负,那么0000 - 0111,0 - 7;1111 - 1000,应该是-0? - -7?这里是直接根据二进制取反(0和1互换)来表示,只是-0 和 +0 是不是有一点点奇怪,一个0有两种表示的同时,还浪费了一个表示数。我们看下一种表示方法
要避免两个0的出现,负数则需要添加一个表示数,0、1.. 7、-8、-7..-1,二进制表示,0111 - 7,1000 - -8,1001 - -7,1111 - -1,我们再单独来看1和-1,0001 - 1,1111 - -1,二进制表示上 1 = -(-1) + 1。所以,你知道原码、反码和补码在描述什么了么

减法

加法规则 1 + 1 = 2,01 + 01 = 10。后面介绍加法器实现的时候,就会发现,加法是很容易通过基础逻辑门实现的,稍微提前一点,当前位是两个数的或运算结果,而进位则是两个数的与运算结果。那减法怎么办,01 - 01 = 00, 10 - 01 = 01,加法需要考虑的只是这一位是否需要向下一位进位,减法则要考虑借位了,那要借多少位呢,是不是有点复杂。当思考陷入僵局时,换一种思维吧
1 - 1 = 1 + (-1) = 0,10 - 01 = 0010 + (-0001)= 0010 + (-1 1111)= 0001.这里看着就有点绕了,为什么一会二进制,一会十进制的,这里是想借十进制的运算规则 减去一个数等于加上一个数的相反数,注意这里的相反数是十进制的取反,而不是二进制的取反。十进制的取反和二进制的取反是不一样的,因为我们没有-0的概念,-8多占了一位,所以十进制的取反 = 二进制的取反之后再加1
以前上学时候背诵的啊,负数的原码是1xxx,xxx是其正数时二进制表示形式,1表示负数;负数的反码等于原码的标识位不变,其余位取反;负数补码等于负数的反码+1,负数的补码常用来进行减法运算,那么你现在明白为什么了吧
打开自带计算器-程序员模式,验证一下

计算 5 - 3,假设第四位为标志位
5 + (-3)= 0101 + (!1011 + 1) = 0101 + (1100 + 1) = 0101 + 1101 = 0010 = 2

逐步实现ALU

半加器

这个为什么叫半加器呢,因为两个半加器可以合成全加器

/**
 * Computes the sum of two bits.
 */
CHIP HalfAdder {
    IN a, b;    // 1-bit inputs
    OUT sum,    // Right bit of a + b 
        carry;  // Left bit of a + b

    PARTS:
    And(a=a,b=b,out=carry);
    Xor(a=a,b=b,out=sum);
}

全加器

半加器的功能对于实现加法是不够的,因为加法是需要三个入参的,即 两个输入数和一个来自低位的进位(初始进位为0)。那么把两个半加器合起来就好了啊

/**
 * Computes the sum of three bits.
 */
CHIP FullAdder {
    IN a, b, c;  // 1-bit inputs
    OUT sum,     // Right bit of a + b + c
        carry;   // Left bit of a + b + c

    PARTS:
    HalfAdder(a=a,b=b,sum=sum1,carry=carry1);
    HalfAdder(a=sum1,b=c,sum=sum,carry=carry2);
    Or(a=carry1,b=carry2,out=carry);
}

16位加法

这里会不会恍惚一下,我们可以实现16位加法了!实现思路也很简单,一位一位的加起来就好了

/**
 * 16-bit adder: Adds two 16-bit two's complement values.
 * The most significant carry bit is ignored.
 */
CHIP Add16 {
    IN a[16], b[16];
    OUT out[16];

    PARTS:
    HalfAdder (a = a[0], b = b[0], sum = out[0], carry = carry0);
    FullAdder (a = carry0, b = a[1], c = b[1], sum = out[1], carry = carry1);
    FullAdder (a = carry1, b = a[2], c = b[2], sum = out[2], carry = carry2);
    FullAdder (a = carry2, b = a[3], c = b[3], sum = out[3], carry = carry3);
    FullAdder (a = carry3, b = a[4], c = b[4], sum = out[4], carry = carry4);
    FullAdder (a = carry4, b = a[5], c = b[5], sum = out[5], carry = carry5);
    FullAdder (a = carry5, b = a[6], c = b[6], sum = out[6], carry = carry6);
    FullAdder (a = carry6, b = a[7], c = b[7], sum = out[7], carry = carry7);
    FullAdder (a = carry7, b = a[8], c = b[8], sum = out[8], carry = carry8);
    FullAdder (a = carry8, b = a[9], c = b[9], sum = out[9], carry = carry9);
    FullAdder (a = carry9, b = a[10], c = b[10], sum = out[10], carry = carry10);
    FullAdder (a = carry10, b = a[11], c = b[11], sum = out[11], carry = carry11);
    FullAdder (a = carry11, b = a[12], c = b[12], sum = out[12], carry = carry12);
    FullAdder (a = carry12, b = a[13], c = b[13], sum = out[13], carry = carry13);
    FullAdder (a = carry13, b = a[14], c = b[14], sum = out[14], carry = carry14);
    FullAdder (a = carry14, b = a[15], c = b[15], sum = out[15], carry = carry15);
}

自增器

实现++1,单独列出来,我想这个很有用。执行程序时,程序指针大部分情况就是+1 +1的

/**
 * 16-bit incrementer:
 * out = in + 1
 */
CHIP Inc16 {
    IN in[16];
    OUT out[16];

    PARTS:
    Add16(a=in,b[0]=true,b[1..15]=false,out=out);
}

ALU

我们终于来到了这一步,不过先调整一下期望值,这里的ALU只实现加法减法、与或非和置零基础运算。这是怎么实现的呢,人为设置了多个开关位,不同的设置来达到目的

/**
 * ALU (Arithmetic Logic Unit):
 * Computes out = one of the following functions:
 *                0, 1, -1,
 *                x, y, !x, !y, -x, -y,
 *                x + 1, y + 1, x - 1, y - 1,
 *                x + y, x - y, y - x,
 *                x & y, x | y
 * on the 16-bit inputs x, y,
 * according to the input bits zx, nx, zy, ny, f, no.
 * In addition, computes the two output bits:
 * if (out == 0) zr = 1, else zr = 0
 * if (out < 0)  ng = 1, else ng = 0
 */
// Implementation: Manipulates the x and y inputs
// and operates on the resulting values, as follows:
// if (zx == 1) sets x = 0        // 16-bit constant
// if (nx == 1) sets x = !x       // bitwise not
// if (zy == 1) sets y = 0        // 16-bit constant
// if (ny == 1) sets y = !y       // bitwise not
// if (f == 1)  sets out = x + y  // integer 2's complement addition
// if (f == 0)  sets out = x & y  // bitwise and
// if (no == 1) sets out = !out   // bitwise not

CHIP ALU {
    IN  
        x[16], y[16],  // 16-bit inputs        
        zx, // zero the x input?
        nx, // negate the x input?
        zy, // zero the y input?
        ny, // negate the y input?
        f,  // compute (out = x + y) or (out = x & y)?
        no; // negate the out output?
    OUT 
        out[16], // 16-bit output
        zr,      // if (out == 0) equals 1, else 0
        ng;      // if (out < 0)  equals 1, else 0

    PARTS:
    Mux16(a=x,b=false,sel=zx,out=x1);
    Not16(in=x1,out=notx1);
    Mux16(a=x1,b=notx1,sel=nx,out=x2);

    Mux16(a=y,b=false,sel=zy,out=y1);
    Not16(in=y1,out=noty1);
    Mux16(a=y1,b=noty1,sel=ny,out=y2);

    Add16(a=x2,b=y2,out=out1);
    And16(a=x2,b=y2,out=out2);
    Mux16(a=out2,b=out1,sel=f,out=out3);

    Not16(in=out3,out=notout3);
    Mux16(a=out3,b=notout3,sel=no,out=out,out[0..7]=out4,out[8..15]=out5,out[15]=ng);
    Or8Way(in=out4,out=or1);
    Or8Way(in=out5,out=or2);
    Or(a=or1,b=or2,out=or3);
    Mux(a=true,b=false,sel=or3,out=zr);
}

这里实现上逐一利用选择器设置合适的值,难点应该是最后两个标志位的设值,如果不知道HDL可以这样表述(最后Mux16这里),还是要多写很多代码的,而且过程中会收到无数次编辑器的提示
可是这怎么用呢。具体的执行逻辑见下表,选一个x-y介绍。更多的,也可以依样推导出来,这里最难的是怎么想到设置这几个开关位就能达到这个效果的,恰好 这个问题是我们不用深究的

x-y

置1的位置为,x=!x x+y -out,out = -(-x + y) = x - y

总结

这节课程的跳跃有点大,好在用到的都是一些熟悉概念,复杂的点在于文字需求 -> 程序代码 -> HDL规范过程中的思维转换,最后的ALU反而并没有想象中那么复杂

posted @ 2024-03-30 22:54  柠檬水请加冰  阅读(51)  评论(0编辑  收藏  举报